Charge given externally to a capacitor

In summary, a capacitor with capacitance C is initially charged to potential difference V from a cell and then disconnected from it. When a charge Q is given to its positive plate, the potential difference across the capacitor is V+Q/2C, instead of V+Q/C as one might expect. This is because only one plate is charged and there is no induced charge on the disconnected negative plate. To solve this problem formally, one can assign surface charge densities to the plates and use three equations to find the charge densities and E field inside the capacitor.
  • #1
EddiePhys
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Homework Statement



A capacitor of capacitance C is charged to a potential difference v from the cell and then disconnected from it. A charge Q is now given to its positive plate. The potential difference across the capacitor is now?

Homework Equations

The Attempt at a Solution


V is the potential difference. Charge +Q is given and hence -Q is induced on the negative plate.

The potential difference should be V+Q/C
But the answer is V+Q/2C. Why?
 
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  • #2
EddiePhys said:
Charge +Q is given and hence -Q is induced on the negative plate.
There is no -Q induced on the negative plate since according to the problem the capacitor has been disconnected from the circuit; there is no path for charge to move onto or off of the negative plate. So there is only a +Q charge on the positive plate.
EddiePhys said:
The potential difference should be V+Q/C
But the answer is V+Q/2C. Why?
The formula ##V=Q/C## assumes equal and opposite charge on both plates. Since only one plate was charged here, it is only half of this value ##Q/2C##.
 
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  • #3
You can solve this formally as follows:
Assign surface charge densities σ1, σ2, -σ2 and σ3 to the two plates. The surfaces 2 and -2 face each other inside the capacitor. Make sure you understand why the inside surface charge densities have to be equal and opposite. Assume unity area. then you can write 3 equations in 3 unknowns and solve for all three surface charge densities.
Two of the equations are simply charge summation. For the third I'll give you a hint: what must the 3rd equation be in order to make the E field zero inside either plate?
Once you have surface charge densities inside the capacitor it should be straightforward to find the E field and integrate, with and without the extra charge.

PS the given answer is what I got.
 
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Related to Charge given externally to a capacitor

1. What is the definition of "charge given externally to a capacitor"?

Charge given externally to a capacitor refers to the process of adding or removing electric charge to or from a capacitor through an external source, such as a battery or power supply.

2. How does charge affect the behavior of a capacitor?

The amount of charge stored in a capacitor determines its capacitance, which is the ability to store electric charge. The higher the charge, the higher the capacitance, and the greater the potential energy stored in the capacitor.

3. Can the charge on a capacitor be changed by manipulating its physical components?

Yes, the charge on a capacitor can be changed by altering its physical components, such as the distance between its plates or the type of dielectric material used. These changes can affect the capacitance and therefore the amount of charge the capacitor can hold.

4. What happens to the charge on a capacitor when it is connected to a circuit?

When a capacitor is connected to a circuit, the charge on the capacitor begins to flow through the circuit, creating a current. As the charge is used up, the voltage across the capacitor decreases until it reaches equilibrium with the voltage of the rest of the circuit.

5. How can the charge on a capacitor be calculated?

The charge on a capacitor can be calculated using the formula Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor. Additionally, the charge can be determined by integrating the current over time, as Q = ∫I(t)dt.

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